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Math Help - Absolute Minimum and Maximum

  1. #1
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    Absolute Minimum and Maximum

    Find the Max and Min values of f on the interval [1/2, 2]

    for the equation:

    f(x) = x - lnx



    I am stuck on this, I tried to find the derivative but I can't figure it out. Any help would be greatly appreciated. Thanks.
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  2. #2
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    Quote Originally Posted by Amybee View Post
    Find the Max and Min values of f on the interval [1/2, 2]

    for the equation:

    f(x) = x - lnx



    I am stuck on this, I tried to find the derivative but I can't figure it out. Any help would be greatly appreciated. Thanks.
    The derivative of the function is just f'(x)=1-\frac{1}{x}. To find the max and mins, set this equal to zero and solve.

    1-\frac{1}{x}=0

    These are the cricitcal points. The only point seems to be at x=1. This is in the given interval, so just find (1,f(1)). If look at the value of the derivative just after x=1, for example f'(2)=\frac{1}{2}, it's clear that the point (1,f(1)) is a minimum. Google the "first derivative test" if you're not familiar with this.

    Edit: the point I made is actually incorrect if you read string6bean1977's reply. I should have used the "Closed Interval Method." However, the point (1,f(1)) is still a minimum.
    Last edited by adkinsjr; November 3rd 2009 at 01:34 PM.
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  3. #3
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    I agree with adkinsjr, but would like to add that the values of x that satisfy f`(x)=0 must then be tested in f(x) to ensure continuity. The zeroes of f`(x) don't automatically make them "c" values. (critical) Additionally, undefined values, such as 0 in the denominator, may also qualify as critical numbers. Such as x=0, in this equation, makes it undefined. However, f(0) doesn't exist either (ln 0), so, in this case, 0 is NOT a "c" value.

    Good luck!

    BB
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  4. #4
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    Quote Originally Posted by string6bean1977 View Post
    I agree with adkinsjr, but would like to add that the values of x that satisfy f`(x)=0 must then be tested in f(x) to ensure continuity. The zeroes of f`(x) don't automatically make them "c" values. (critical) Additionally, undefined values, such as 0 in the denominator, may also qualify as critical numbers. Such as x=0, in this equation, makes it undefined. However, f(0) doesn't exist either (ln 0), so, in this case, 0 is NOT a "c" value.

    Good luck!

    BB
    oh, your right! I didn't notice that. It's a good thing someone mentioned that.
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